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Dawny521 · Jun 23, 2009, 05:02 PM

Can you help me with this problem please I don't even know where to begin with what's in parentheses are fractions

-28x^3y^(-1/4)
4x^(-1/5)y =

Perito · Jun 23, 2009, 05:58 PM

Originally Posted by Dawny521

-28x^3y^(-1/4)
4x^(-1/5)y =

Remember that $x^{-a} = \frac {1}{x^a}$

Now I hope I understood your problem correctly. It was not trivial to figure out. Next time, try using parentheses and "/" for division and ^ for exponentiation.

[-28x^3 y^(-1/4)] / [4x^(-1/5) y]

or, better, learn how to use LaTeX ;)

Post on using LaTeX

$\frac {-28x^3y^{(-1/4)}}{4x^{(-1/5)}y} =$

$\frac {-28x^3 \frac {1}{y^{(1/4)}}}{4 \frac {1}{x^{(1/5)}} y } = \frac {-28x^3 x^{(1/5)}}{4y^{(1/4)} y} = \frac {-28 x^{(16/5)}}{ 4 y^{(5/4)}} = \frac {-7 x^{\frac {16}{5}}}{ y^{\frac 54}}$

jcaron2 · Jun 23, 2009, 07:48 PM

$\frac{-28x^3y^{-\frac1 4}}{4x^{-\frac1 5}y}\;=\\<br /> \frac{-28x^3x^{\frac1 5}}{4y^{\frac1 4}y}\;=\\<br /> \frac{-7x^{\frac{16} 5}}{y^{\frac5 4}}<br />$

Perito · Jun 23, 2009, 07:50 PM

Thanks. Did you drop 4 in the denominator? I can't see where it went.

LaTeX gets ugly with all of those fractions. Doesn't it?

jcaron2 · Jun 23, 2009, 08:13 PM

LOL! Yes, I did (and yes, it does). How many engineers does it take to solve a simple algebra problem?

Thanks to the edit button, all evidence of our ineptitude has disappeared.